Question Number 13658 by Tinkutara last updated on 22/May/17
![Prove that cos^2 x + cos^2 3x + cos^2 5x + ... to n terms = (1/2)[n + ((sin4nx)/(2sin2x))]](https://www.tinkutara.com/question/Q13658.png)
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{cos}^{\mathrm{2}} {x}\:+\:\mathrm{cos}^{\mathrm{2}} \mathrm{3}{x}\:+\:\mathrm{cos}^{\mathrm{2}} \mathrm{5}{x}\:+\:…\:\mathrm{to}\:{n}\:\mathrm{terms} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left[{n}\:+\:\frac{\mathrm{sin4}{nx}}{\mathrm{2sin2}{x}}\right] \\ $$
Answered by ajfour last updated on 22/May/17
![2S=(1+cos 2x)+(1+cos 6x) + (1+cos 10z)+....+(1+cos [(4n−2)x]) 2S=n+(1/(2sin 2x)){ 2cos 2xsin 2x+2cos 6xsin 2x+2cos 10xsin 2x+.... .....+2cos [4(n−2)x]sin 2x } 2S=n+(1/(2sin 2x)){sin 4x+ sin 8x−sin 4x + sin 12x−sin 8x +......... − ...... +sin [(4n−4)x]−sin [(4n−8)x] +sin 4nx−sin [(4n−4)x] } 2S=n+((sin 4nx)/(2sin 2x)) . S=(1/2)[n+((sin 4nx)/(2sin 2x))] .](https://www.tinkutara.com/question/Q13667.png)
$$\mathrm{2}{S}=\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\right)+\left(\mathrm{1}+\mathrm{cos}\:\mathrm{6}{x}\right) \\ $$$$\:\:\:\:\:+\:\left(\mathrm{1}+\mathrm{cos}\:\mathrm{10}{z}\right)+….+\left(\mathrm{1}+\mathrm{cos}\:\left[\left(\mathrm{4}{n}−\mathrm{2}\right){x}\right]\right) \\ $$$$\mathrm{2}{S}={n}+\frac{\mathrm{1}}{\mathrm{2sin}\:\mathrm{2}{x}}\left\{\:\mathrm{2cos}\:\mathrm{2}{x}\mathrm{sin}\:\mathrm{2}{x}+\mathrm{2cos}\:\mathrm{6}{x}\mathrm{sin}\:\mathrm{2}{x}+\mathrm{2cos}\:\mathrm{10}{x}\mathrm{sin}\:\mathrm{2}{x}+….\right. \\ $$$$\left.\:\:\:\:\:\:\:…..+\mathrm{2cos}\:\left[\mathrm{4}\left({n}−\mathrm{2}\right){x}\right]\mathrm{sin}\:\mathrm{2}{x}\:\right\} \\ $$$$\mathrm{2}{S}={n}+\frac{\mathrm{1}}{\mathrm{2sin}\:\mathrm{2}{x}}\left\{\mathrm{sin}\:\mathrm{4}{x}+\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{sin}\:\mathrm{8}{x}−\mathrm{sin}\:\mathrm{4}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\mathrm{sin}\:\mathrm{12}{x}−\mathrm{sin}\:\mathrm{8}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+………\:\:−\:\:\:…… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{sin}\:\left[\left(\mathrm{4}{n}−\mathrm{4}\right){x}\right]−\mathrm{sin}\:\left[\left(\mathrm{4}{n}−\mathrm{8}\right){x}\right] \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{sin}\:\mathrm{4}{nx}−\mathrm{sin}\:\left[\left(\mathrm{4}{n}−\mathrm{4}\right){x}\right]\:\right\} \\ $$$$\mathrm{2}{S}={n}+\frac{\mathrm{sin}\:\mathrm{4}{nx}}{\mathrm{2sin}\:\mathrm{2}{x}}\:. \\ $$$$\boldsymbol{{S}}=\frac{\mathrm{1}}{\mathrm{2}}\left[{n}+\frac{\mathrm{sin}\:\mathrm{4}{nx}}{\mathrm{2sin}\:\mathrm{2}{x}}\right]\:\:. \\ $$
Commented by Tinkutara last updated on 23/May/17
$$\mathrm{Thanks}! \\ $$